Monotone Functions

• November 9th 2010, 09:59 AM
zebra2147
Monotone Functions
I have attempted this proof and would appreciate any feedback. Thanks.

Let $I$ be some interval. If $f:I\rightarrow \mathbb{R}$ is increasing and $a\in I$, prove that $f(a)$ is an upper bound for $\{f(x)|x and $lim_{x\nearrow a}f(x)=sup\{f(x)|x.

Proof: Let $f$ be an increasing function. Then, by definition, for all $x,y\in I, x. Then, let $x for $a\in I$ and all $x\in I$. Thus, since $x for all $a,x\in I$. Hence, $a$ is an upper bound for $\{f(x)|x.

Then, since $x for all $x,a\in I$, then $a$ must be the least upper bound for $\{f(x)|x. Hence, $lim_{x\nearrow a}f(x)=sup\{f(x)|x.
• November 9th 2010, 10:52 AM
Plato
You have some confused notation here.
You have $a$ in two different ways.
What you want is to note that $f(a)$ is an upper bound for $\{f(x):x.
Now consider $A=\sup\{f(x):x so that $A\le f(a)$.
Then show that $\displaystyle \lim _{x \to a^ - } f(x) = A$
• November 9th 2010, 11:09 AM
emakarov
Major remarks.

Quote:

Hence, $a$ is an upper bound for $\{f(x)|x.
Should be f(a), not a.

The second claim is not really proved.
Quote:

Then, since $x for all $x,a\in I$, then $a$ must be the least upper bound for $\{f(x)|x.
Even the claim that f(a) (rather than a) is the least upper bound for $\{f(x)\mid x is incorrect. Consider

$f(x)=
\begin{cases}
x & x x+1 & x\ge a
\end{cases}
$

Then $\sup\{f(x)\mid x, but $f(a)=a+1$.

Even if $f(a)=\sup\{f(x)\mid x, how does it help prove the following?
Quote:

Hence, $lim_{x\nearrow a}f(x)=sup\{f(x)|x.
After all, $\lim_{x\nearrow a}f(x)=f(a)$ only if f is left-continuous at a. In my opinion, there is at least not enough details here that are expected from a proof of such a simple fact.

Minor remarks.

Quote:

Then, let $x for $a\in I$ and all $x\in I$.
You can't expect that x < a for all x in I. Since a is already fixed in the problem statement, this phrase should say, "Let x < a for some x in I".

Quote:

Thus, since $x for all $a,x\in I$.
Again, it is not the case that f(x) < f(a) holds for all a, x in I.
• November 10th 2010, 05:53 AM
zebra2147
Alright, so from what I gather, most of the first part was ok besides the fact that $f(a)$ is the upper bound and not $a$.

As for the second part...
Let $A=\sup\{f(x):x and $A\leq f(a)$. Then,
since $f(a)$ is an upper bound for $\{f(x)|x and f is increasing then there is no $x>a$ such that $f(x)>f(a)$. Thus, $lim_{x\rightarrow a^-}f(x)\leq f(a)$. If we let this limit be called $A$ then we have that $lim_{x\rightarrow a^-}f(x)=A$ where $A\leq f(a)$. Hence, we have that $lim_{x\nearrow a}f(x)=sup\{f(x)|x.

I'm not sure this is very strong but its my second attempt.
• November 10th 2010, 07:34 AM
emakarov
Sorry, too many flaws. The second part of the problem -- proving $\lim_{x\to a^-}f(x)=\sup\{f(x)\mid x -- has nothing to do with f(a). In the example I gave above, f(a) is different from $\sup\{f(x)\mid x.

Quote:

Let $A=\sup\{f(x):x and $A\leq f(a)$.
Are you assuming $A\leq f(a)$? In this case, since this assumption is not in the problem statement, you later need to consider the possibility that $A>f(a)$. However, I thought that you have proved this in part 1. Therefore, you should say, "Let $A=\sup\{f(x):x. We have already shown that $A\leq f(a)$." (To repeat, the second sentence is irrelevant.)

Quote:

Then, since $f(a)$ is an upper bound for $\{f(x)|x and f is increasing then there is no $x>a$ such that $f(x)>f(a)$.
What happens to the right of $a$ is irrelevant to the problem.

Quote:

Thus, $lim_{x\rightarrow a^-}f(x)\leq f(a)$.
I don't see how this follows from the previous discussion.

Quote:

If we let this limit be called $A$
This is redefining A.

Quote:

then we have that $\lim_{x\rightarrow a^-}f(x)=A$ where $A\leq f(a)$. Hence, we have that $\lim_{x\nearrow a}f(x)=sup\{f(x)|x.
It looks like this conclusion is based on confusing two meanings of A.

Let us define $A=\sup\{f(x)\mid x, as you have done. Here is how I would prove that $\lim_{x\to a^-}f(x)=A$. We show that for every $\varepsilon>0$ there exists a $\delta>0$ such that when $a-\delta and $x\in I$ we have $A-\varepsilon. Note that the second inequality, i.e., $f(x)\le A$, holds by the definition of $\sup$.

Let $\varepsilon$ be given. Also by the definition of $\sup$, $A-\varepsilon$ is not an upper bound of $\{f(x)\mid x; therefore, there exists an $x_0 such that $f(x_0)>A-\varepsilon$. But then, since f is increasing, for every $x$ such that $x_0 we have $A-\varepsilon. Therefore, we can take $\delta=a-x_0$.
• November 10th 2010, 03:58 PM
Plato
A quasi-continuous function is one for which the right and left hand limits exist at each point. It is a standard theorem in analysis that any monotonic function is quasi-continuous. That is the essence of this problem. You are asked to show that the limit on the left exist at each point.